[ Problem View ]
Gauss's Law in 3, 2, and 1 Dimension
Gauss's law relates the electric flux Phi_E through a closed surface to the total charge q_encl enclosed by the surface:

\Phi_E = \oint \vec{E}\cdot d\vec{A} = \frac{q_{\rm encl}}{\epsilon_0}.

You can use Gauss's law to determine the charge enclosed inside a closed surface on which the electric field is known. However, Gauss's law is most frequently used to determine the electric field from a symmetric charge distribution.

The simplest case in which Gauss's law can be used to determine the electric field is that in which the charge is localized at a point, a line, or a plane. When the charge is localized at a point, so that the electric field radiates in three-dimensional space, the Gaussian surface is a sphere, and computations can be done in spherical coordinates. Now consider extending all elements of the problem (charge, Gaussian surface, boundary conditions) infinitely along some direction, say along the z axis. In this case, the point has been extended to a line, namely, the z axis, and the resulting electric field has cylindrical symmetry. Consequently, the problem reduces to two dimensions, since the field varies only with x and y, or with r and theta in cylindrical coordinates. A one-dimensional problem may be achieved by extending the problem uniformly in two directions. In this case, the point is extended to a plane, and consequently, it has planar symmetry.

 

Consider a point charge q in three-dimensional space. Symmetry requires the electric field to point directly away from the charge in all directions. To find E(r), the magnitude of the field at distance r from the charge, the logical Gaussian surface is a sphere centered at the charge. The electric field is normal to this surface, so the dot product of the electric field and an infinitesimal surface element involves \cos(0)=1. The flux integral is therefore reduced to \int E(r)\, dA=E(r) A(r), where E(r) is the magnitude of the electric field on the Gaussian surface, and A(r) is the area of the surface.

Part A
Determine the magnitude E(r) by applying Gauss's law.
Part A.1  

Part not displayed

Express E(r) in terms of some or all of the variables/constants q, r, and epsilon_0.
ANSWER:
  E(r) =  \frac{1}{4{\pi}{\epsilon}_{0}}\frac{q}{r^{2}} 
 
Now consider the case that the charge has been extended along the z axis. This is generally called a line charge. The usual variable for a line charge density (charge per unit length) is lambda, and it has units (in the SI system) of coulombs per meter.
Part B
By symmetry, the electric field must point radially outward from the wire at each point; that is, the field lines lie in planes perpendicular to the wire. In solving for the magnitude of the radial electric field E(r) produced by a line charge with charge density lambda, one should use a cylindrical Gaussian surface whose axis is the line charge. The length of the cylindrical surface L should cancel out of the expression for E(r). Apply Gauss's law to this situation to find an expression for E(r).
Part B.1  

Part not displayed

Part B.2  

Part not displayed

Express E(r) in terms of some or all of the variables lambda, r, and any needed constants.
ANSWER:
  E(r) =  \frac{1}{2{\pi}{\epsilon}_{0}}\frac{{\lambda}}{r} 
 
Now consider the case with one effective direction. In order to make a problem effectively one-dimensional, it is necessary to extend a charge to infinity along two orthogonal axes, conventionally taken to be x and y. When the charge is extended to infinity in the xy plane (so that by symmetry, the electric field will be directed in the z direction and depend only on z), the charge distribution is sometimes called a sheet charge. The symbol usually used for two-dimensional charge density is either sigma, or eta. In this problem we will use eta. eta has units of coulombs per meter squared.
Part C
In solving for the magnitude of the electric field E_vec(z) produced by a sheet charge with charge density eta, one may use a Gaussian surface in the shape of a rectangular box two of whose faces lie a distance z above and below the sheet of charge. The area A of these faces must then be calculated; they will cancel out of the expression for E(z) in the end. The result of applying Gauss's law to this situation then gives an expression for E(z) for both z>0 and z<0.
Part C.1  

Part not displayed

Part C.2  

Part not displayed

Express E(z) for z>0 in terms of some or all of the variables/constants eta, z, and epsilon_0.
ANSWER:
  E(z) =  \frac{1}{2{\epsilon}_{0}}{\eta} 

In this problem, the electric field from a distribution of charge in 3, 2, and 1 dimension has been found using Gauss's law. The most noteworthy feature of the three solutions is that in each case, there is a different relation of the field strength to the distance from the source of charge. In each case, the field strength varies inversely as an integral power of the distance r from the charge. In the case of a point charge (spherical symmetry, field in three dimensions), the field strength varies as r^{-2}. In the case of a line charge (cylindrical symmetry, field in two dimensions), the field strength varies as r^{-1}. Finally, in the case of a sheet charge (planar symmetry, field in one dimension), the field varies as r^0=1; that is, the strength of the field is independent of the distance from the sheet!

If you visualize the electric field using field lines, this result shows that as the number of directions in which the electric field can point is reduced, the field lines have one dimension fewer in which to to spread out, and the field therefore falls off less rapidly with distance. In a one-dimensional problem (sheet charge), the extension of the charge in the xy plane means that all field lines are parallel to the z axis, and so the field strength does not change with distance. Such a situation, of course, is impossible in the real world: In reality, the planar charge is not infinite, so the field will in fact fall off over long distances.

 [ Print ]